Producting Limits...

$\underset { n\rightarrow \infty }{ lim } \prod _{ r=3 }^{ n }{ \frac { { r }^{ 3 }-8 }{ { r }^{ 3 }+8 } }$

$=\quad \underset { n\rightarrow \infty }{ lim } \left( \frac { { 3 }^{ 3 }-8 }{ { 3 }^{ 3 }+8 } \right) \left( \frac { { 4 }^{ 3 }-8 }{ 4^{ 3 }+8 } \right) ........\left( \frac { { n }^{ 3 }-8 }{ n^{ 3 }+8 } \right)$

$=\quad \underset { n\rightarrow \infty }{ lim } \left( \frac { 3-2 }{ 3+2 } .\frac { { 3 }^{ 2 }+4+2(3) }{ { 3 }^{ 2 }+4-2(3) } \right) \left( \frac { 4-2 }{ 4+2 } .\frac { { 4 }^{ 2 }+4+2(4) }{ 4^{ 2 }+4-2(4) } \right) .........\left( \frac { n-2 }{ n+2 } .\frac { n^{ 2 }+4+2(n) }{ { n }^{ 2 }+4-2(n) } \right)$

$=\quad \underset { n\rightarrow \infty }{ lim } \left\{ \frac { 3-2 }{ 3+2 } .\frac { 4-2 }{ 4+2 } .\frac { 5-2 }{ 5+2 } .......\frac { n-2 }{ n+2 } \right\} \left\{ \frac { { 3 }^{ 2 }+4+2(3) }{ { 3 }^{ 2 }+4-2(3) } .\frac { { 4 }^{ 2 }+4+2(4) }{ 4^{ 2 }+4-2(4) } .......\frac { { n }^{ 2 }+4+2(n) }{ { n }^{ 2 }+4-2(n) } \right\}$

$=\quad \underset { n\rightarrow \infty }{ lim } \left\{ \frac { 1.2.3.4.5.6.... }{ 5.6.7.8.... } \right\} \left\{ \frac { 19.28.39.52.63....... }{ 7.12.19.28.39.42.63..... } \right\}$

$=\quad \underset { n\rightarrow \infty }{ lim } \frac { \left( { n }^{ 2 }+5 \right) \left( { n }^{ 2 }+2n+4 \right) \left( 1.2.3.4 \right) }{ \left( { n }-1 \right) n\left( { n }+1 \right) \left( { n }+2 \right) } \left( \frac { 1 }{ 7.12 } \right)$

$= \quad \boxed{\frac{2}{7}}$

i'm going to use a simple telescoping sum to determine the answer. first off, $S=\prod_{k=3}^\infty (\dfrac{k^3-8}{k^3+8})$ taking natural logarithm on both side $\ln (S)=\sum_{k=3}^\infty (\ln(\dfrac{(k-2)(k^2+2k+4)}{(k+2)(k^2-2k+4)}))$ expanding by rules of logarithms, $\ln S=\sum_{k=3}^\infty (\ln(k-2)-\ln(k+2)+\ln(k^2+2k+4)-\ln(k^2-2k+4))$ expanding, $\ln S =\sum_{k=3}^\infty (\ln(k-2)-\ln(k+2))-\sum_{k=3}^\infty (\ln(k^2-2k+4)-\ln(k^2+2k+4))$ both are telescoping series,and solving them we get. $\ln S= (\ln (1)+\ln (2)+\ln (3)+\ln (4))-(\ln (7)+\ln (12))=\ln(\dfrac{2}{7})$ hence $S=\dfrac{2}{7}, 2+7=\boxed{9}$

$\prod_{r=3}^{\infty}\big(\frac{r^3-8}{r^3+8}\big)=\prod_{r=3}^{\infty}\frac{(r-2)(r^2+2r+4)}{(r+2)(r^2-2r+4)}$

$=\frac{\prod_{r=3}^{\infty}(r-2)\prod_{r=3}^{\infty}(r^2+2r+4)}{\prod_{r=3}^{\infty}(r+2)\prod_{r=3}^{\infty}(r^2-2r+4)}$

$=\frac{1\cdot 2\cdot 3\cdot 4 \prod_{r=7}^{\infty}(r-2)\prod_{r=3}^{\infty}(r^2+2r+4)} {7 \cdot 12 \prod_{r=3}^{\infty}(r+2)\prod_{r=5}^{\infty}(r^2-2r+4)}$

By changing the limits, we get

$\prod_{r=3}^{\infty}\big(\frac{r^3-8}{r^3+8}\big)=\frac{1\cdot 2\cdot 3\cdot 4 \prod_{r=3}^{\infty}(r+2)\prod_{r=3}^{\infty}(r^2+2r+4)} {7 \cdot 12 \prod_{r=3}^{\infty}(r+2)\prod_{r=3}^{\infty}(r^2+2r+4)}=\large\boxed{\frac{2}{7}}$

Product of [(r - 2)/ (r + 2)][(r^2 + 2 x + 4)/ (r^2 - 2 x + 4)]

= (1)(2)(3)(4)/ [(7)(12)]

= 2/ 7

2 + 7 = 9